Теория: 15 Умножение и деление буквенных выражений, содержащих степень с рациональным показателем

Получаем:

\(\displaystyle \frac{a^{\frac{2}{7}}-2a^{\frac{1}{7}}b^{\frac{1}{5}}+b^{\frac{2}{5}}}{4b^{\frac{2}{5}}+12b^{\frac{1}{5}}+9}\cdot \frac{8a^{\frac{2}{7}}b^{\frac{1}{5}}+12a^{\frac{2}{7}}}{4a^{\frac{3}{7}}-4a^{\frac{2}{7}}b^{\frac{1}{5}}}=\frac{(a^{\frac{2}{7}}-2a^{\frac{1}{7}}b^{\frac{1}{5}}+b^{\frac{2}{5}})\cdot (8a^{\frac{2}{7}}b^{\frac{1}{5}}+12a^{\frac{2}{7}})}{(4b^{\frac{2}{5}}+12b^{\frac{1}{5}}+9)\cdot (4a^{\frac{3}{7}}-4a^{\frac{2}{7}}b^{\frac{1}{5}})}{\small .}\)

Разложим выражения в числителе и знаменателе на множители:

• \(\displaystyle a^{\frac{2}{7}}-2a^{\frac{1}{7}}b^{\frac{1}{5}}+b^{\frac{2}{5}}=a^{\frac{2}{7}}-2\cdot a^{\frac{1}{7}}\cdot b^{\frac{1}{5}}+b^{\frac{2}{5}}=(a^{\frac{1}{7}}-b^{\frac{1}{5}})^2\small,\)
• \(\displaystyle 8a^{\frac{2}{7}}b^{\frac{1}{5}}+12a^{\frac{2}{7}}=4a^{\frac{2}{7}}(2b^{\frac{1}{5}}+3)\small,\)
• \(\displaystyle 4b^{\frac{2}{5}}+12b^{\frac{1}{5}}+9=(2b^{\frac{1}{5}})^2+2\cdot2b^{\frac{1}{5}}\cdot3+3^2= (2b^{\frac{1}{5}}+3)^2\small,\)
• \(\displaystyle 4a^{\frac{3}{7}}-4a^{\frac{2}{7}}b^{\frac{1}{5}}=4a^{\frac{2}{7}}(a^{\frac{1}{7}}-b^{\frac{1}{5}}){\small .}\)

Подставляя, получаем:

\(\displaystyle \frac{(a^{\frac{2}{7}}-2a^{\frac{1}{7}}b^{\frac{1}{5}}+b^{\frac{2}{5}})\cdot (8a^{\frac{2}{7}}b^{\frac{1}{5}}+12a^{\frac{2}{7}})}{(4b^{\frac{2}{5}}+12b^{\frac{1}{5}}+9)\cdot (4a^{\frac{3}{7}}-4a^{\frac{2}{7}}b^{\frac{1}{5}})}= \frac{(a^{\frac{1}{7}}-b^{\frac{1}{5}})^2 \cdot 4a^{\frac{2}{7}}(2b^{\frac{1}{5}}+3)}{{(2b^{\frac{1}{5}}+3)^2}\cdot 4a^{\frac{2}{7}}(a^{\frac{1}{7}}-b^{\frac{1}{5}})} =\)

\(\displaystyle = \frac{4a^{\frac{2}{7}}(a^{\frac{1}{7}}-b^{\frac{1}{5}})^2 (2b^{\frac{1}{5}}+3)}{{4a^{\frac{2}{7}}(2b^{\frac{1}{5}}+3)^2}(a^{\frac{1}{7}}-b^{\frac{1}{5}})} {\small .}\)

Ответ: \(\displaystyle \frac{a^{\frac{1}{7}}-b^{\frac{1}{5}}}{2b^{\frac{1}{5}}+3} \small.\)